Question

(2-Dimensial Collision) A billiard ball with speed v = 2.0m/s approaches an identical stationary one (see figure at the right). The balls bounce off each other elastically, in such a way that the incoming one gets deflected by an angle θ = 35°. a. What are the final speeds of the balls? b. What is the angle, φ, at which the stationary ball is ejected?

Answer 1

Along x direction

$m u=0=mv_f cos \theta+mv_f cos \phi$

$u=v_f cos \theta+v_f cos \phi.........(1)$

Along y direction

$0=mv_fsin \theta+mv_f sin \phi$

$u=v_f cos \theta+v_f cos \phi.........(2)$

$sin \theta= sin \phi$

$\theta= \phi.........(3)$

Substituting (1)

$u=v_f cos \theta+v_f cos \theta$

$u=2v_f cos \theta$

$v_f = \frac{u}{2cos \theta}= \frac{2}{2cos 35}=2.44154 m/s$

As the collision is elastic , $\frac{1}{2}m u^2 =\frac{1}{2} m v^2_f +\frac{1}{2} m v^2_f$

$u^2 =2 v^2_f$

$v_f=\frac{u}{\sqrt{2}}.........(4)$

Substituting (4) in (3)

$v_f=\frac{u}{2 cos \theta}=\frac{u}{\sqrt{2}}$

$2cos \theta=\sqrt{2}$

$cos \theta=\frac{\sqrt{2}}{2}$

$\theta=45^0$